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Question

At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?

Solution

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First, we convert the linear speed units to meters per second. Then, we proceed to calculate the angular velocity with the given diameter (d)\left(d\right). Note that we have substituted the radius (r)\left(r\right) for d/2d/2.

v=500 kmh1000 m1 km1 h3600 s=139 msω=vr=2vd=2(139)60.0=4.63 rads=4.63 rads1 rev2π rad=0.737 revs\begin{align*} v&=500 \text{ } \dfrac{\text{km}}{\text{h}} \cdot \dfrac{1000 \text{ m}}{1 \text{ km}} \cdot \dfrac{1 \text{ h}}{3600\text{ s}} \\ &=139 \text{ } \dfrac{\text{m}}{\text{s}} \\ \omega&=\dfrac{v}{r} \\ &=\dfrac{2v}{d} \\ &=\dfrac{2\left(139\right)}{60.0} \\ &=4.63 \text{ } \dfrac{\text{rad}}{\text{s}} \\ &=4.63 \text{ } \dfrac{\text{rad}}{\text{s}} \cdot \dfrac{1 \text{ rev}}{2\pi \text{ rad}} \\ &=0.737 \text{ } \dfrac{\text{rev}}{\text{s}} \end{align*}

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